\newproblem{lay:7_3_12}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.3.12}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ be a symmetric $n\times n$ matrix, let $M$ and $m$ denote the maximum and minimum values of the quadratic form $\mathbf{x}^TA\mathbf{x}$ subtject to $\|\mathbf{x}\|=1$. Let $\lambda$ be any eigenvalue of
	$A$. Justify that $m\leq \lambda \leq M$. (\textit{Hint}: Find an $\mathbf{x}$ such that $\mathbf{x}^TA\mathbf{x}=\lambda$.)
}{
   % Solution
	Thanks to the Principal Axes Theorem (Theorem 7.2.4) we know that by diagonalizing matrix $A=PDP^T$ we can express the quadratic form $\mathbf{x}^TA\mathbf{x}$ as
	\begin{equation}
		 \mathbf{x}^TA\mathbf{x}=\mathbf{y}^TD\mathbf{y}=\lambda_1y_1^2+...+\lambda_ny_n^2
	\end{equation}
	where we have made a change of variable $\mathbf{x}=P\mathbf{y}$. Let $\lambda$ in the problem be the $i$-th eigenvalue of $A$ and consider a unitary vector with a single 1 at the
	$i$-th positition ($\mathbf{y}=(0,0,...,0,1,0,...,0)$). It is obvious that
	\begin{equation}
		 \lambda=\mathbf{x}^TA\mathbf{x}=\mathbf{y}^TD\mathbf{y}=\lambda_10^2+...+\lambda_{i-1}0^2+\lambda_i1^2+\lambda_{i+1}0^2+...+\lambda_n0^2=\lambda_i
	\end{equation}
	This result is attained for $\mathbf{x}=P\mathbf{y}=\mathbf{u}_i$, that is, the eigenvector associated to the $i$-th eigenvalue.
	
	Additionally, thanks to Theorem 7.3.6, we know that the quadratic form $\mathbf{x}^TA\mathbf{x}$ is bounded between the minimum and maximum eigenvalue of $A$ when $\mathbf{x}$ is constrained to be unitary. Moreover,
	$m=\lambda_{min}$ and $M=\lambda_{max}$. Consequently, since $\lambda_{min}\leq \lambda \leq \lambda_{max}$, we have $m\leq \lambda \leq M$.
	
}
\useproblem{lay:7_3_12}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

